Prove that any similiarity transformation with ratio of similarity $k=\frac{1}{2}$ in three-dimensional space $E^3$ has one fixed point.
It is easy to show that similiarity transformation can't have more then one fixed point (Let $A$ and $B$ are fixed points. Then $|f(A)f(B)| = k|AB|$ or $|AB|=k|AB|$. Contradiction). But how to show that such transformation has at least one fixed point?
Take a point $A$ and suppose $T(A) = A'$. If the fixed point is $C$, then $d(A',C) = (1/2) d(A,C)$. So if we consider all points $X$ with $2 d(X, A') = d(X, A)$, then $X$ must be on this locus. It turns out (which I discovered by doing a little analytic geometry with $A = (1, 0)$ and $A' = (0,0)$ that this locus is a circle with center $Q$, where $Q-A'-A$ and $d(Q, A') = d(A', A)$, i.e., $Q$ is the reflection of $A$ through $A'$. And the radius of the circle is $r = \sqrt(2)d(A, A')$.
Now let $T(Q) = Q'$; then the ray $QQ'$ must pass through $C$, and must be a distance $r$ from $Q$. So you can construct $C$ as the point $r$ units from $Q$ along the ray from $Q$ to $T(Q)$.
This argument shows that if there is a fixed point, it must be at the point I've constructed. I leave it to you to show that this is in fact a fixed point. (Arguing about the distances from $C$ to $A$ and $A'$ should get you 90% of the way there.)
Thanks for asking a really fun question!
(By the way, nothing in the work I did really required the ratio be $1/2$; if you have some other ratio (except $\pm 1$) then an exactly similar argument lets you find the point which must be the center.)